Question

Sử dụng hằng đẳng thức để tìm x:

(x+1)^3 - (x-1)^3 - 6(x-2)(x+1) +3x -2=0

Answer 1

\(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-2\right)\left(x+1\right)+3x-2=0\)

\(x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+6x+12+3x-2=0\)

\(1+1+6x+3x+12-2=0\)

\(9x+12=0\)

\(9x=-12\)

\(x=\frac{-4}{3}\)

Answer 2

\(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-2\right)\left(x+1\right)+3x-2=0\)

\(\Leftrightarrow\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-2\right)\left(x+1\right)+3x=0+2\)

\(\Leftrightarrow\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-2\right)\left(x+1\right)+3x=2\)

\(\Leftrightarrow9x+14=2\)

\(\Leftrightarrow9x=2-14\)

\(\Leftrightarrow9x=-12\)

\(\Leftrightarrow x=\frac{-12}{9}=\frac{-4}{3}\)

\(\Rightarrow x=\frac{-4}{2}\)